Minimum number of colors enough to color the vertices of any graph whose vertex lies on at most $k$-odd...
$begingroup$
I've got no time to solve this problem during the exam as it's the last one.
What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?
But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.
discrete-mathematics graph-theory coloring
edited Jan 6 at 16:59
Namaste
1
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
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add a comment |
$begingroup$
I've got no time to solve this problem during the exam as it's the last one.
What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?
But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.
discrete-mathematics graph-theory coloring
edited Jan 6 at 16:59
Namaste
1
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
$endgroup$
add a comment |
$begingroup$
I've got no time to solve this problem during the exam as it's the last one.
What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?
But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.
discrete-mathematics graph-theory coloring
edited Jan 6 at 16:59
Namaste
1
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
$endgroup$
I've got no time to solve this problem during the exam as it's the last one.
What is the minimum number of colors that would suffice to color a graph so that adjacent nodes get different colors if each node lies on at most $k$-odd cycles?
But I still have no idea about it (even unsure about the answer). Hints will also be appreciated.
discrete-mathematics graph-theory coloring
discrete-mathematics graph-theory coloring
edited Jan 6 at 16:59
Namaste
1
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
edited Jan 6 at 16:59
Namaste
1
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
edited Jan 6 at 16:59
Namaste
1
edited Jan 6 at 16:59
Namaste
1
edited Jan 6 at 16:59
Namaste
1
1
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
asked Jan 6 at 7:13
Oolong milk teaOolong milk tea
787
787
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
Not sure if this is correct:
Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.
So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Not sure if this is correct:
Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.
So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
add a comment |
$begingroup$
Not sure if this is correct:
Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.
So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
$endgroup$
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
add a comment |
$begingroup$
Not sure if this is correct:
Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.
So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
$endgroup$
Not sure if this is correct:
Let $G$ be the graph. Assume that there is a vertex $vin V(G)$ that lie on $k$-odd cycles, otherwise each node could lie on $k+n$ cycles for any $nin mathbb{N}$. And this just says the chromatic number could be anything really.
So choose the node $v in V(G)$ which lie on $k$ odd cycle, removes it. then all $k$ odd cycles are gone. Any node on those cycles cannot be on the $k$ removed cycles in the remaining graph. So if you see another node $v_2$ that is in $k$ odd cycles, then it is not adjacent to $v$. So removes $v_2$ to remove the $k$-odd cycles that $v_2$ is in. Continue this way until stuck. Notice all the removed nodes can be color using the same color so let's just color them with color $k$. The resulting graph has the property that all node are in at most $k-1$ cycles. So as before, pick a vertex $v^2$ that is adjacent to $k-1$ cycles, if there is any. Removes those one by one as before and notice they are all pairwise no adjacent and hence can be color using the same color. Color the removed vertices using color $k-1$. OK then you do this at most $k$ steps to get a no odd cycle graph, which can be colored using two colors as it would be bipartite. These 2 colors together with the $k$ color used, we get a $k+2$ coloring of the graph?
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
edited Jan 6 at 23:44
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
answered Jan 6 at 23:37
nafhgoodnafhgood
1,803422
1,803422
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
add a comment |
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Please explain the part "then it is not adjacent to $v$". How do you know it's not adjacent to $v$?
$endgroup$
– bof
Jan 6 at 23:50
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
$begingroup$
Ah right sorry...I'm actually not sure about it either...
$endgroup$
– nafhgood
Jan 7 at 0:42
add a comment |
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